Kingdom of Black and White（模拟）

http://acm.hdu.edu.cn/showproblem.php?pid=5583

题意： 给你一个只含有01的字符串，然后这个串的权值就是每一段连续的0或1的长度的平方和，然后你可以修改一个数，使得这个数变成0，或者使这个数变成1

然后问你最大权值能为多少。

解法：暴力枚举改变每一个联通块，注意当联通块值为1时，需合并3块联通块。

#include<bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <string>
#include <stdio.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string.h>
#include <vector>
using namespace std;
typedef long long ll ;
#define int ll
#define mod 1000000007
#define gcd(m,n) __gcd(m, n)
#define rep(i , j , n) for(int i = j ; i <= n ; i++)
#define red(i , n , j)  for(int i = n ; i >= j ; i--)
#define ME(x , y) memset(x , y , sizeof(x))
int lcm(int a , int b){return a*b/gcd(a,b);}
//ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;}
//int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;}
//const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len}
#define INF  0x3f3f3f3f
#define PI acos(-1)
#define pii pair<int,int>
#define fi first
#define se second
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define pb push_back
#define mp make_pair
#define all(v) v.begin(),v.end()
#define size(v) (int)(v.size())
#define cin(x) scanf("%lld" , &x);
const int N = 1e7+9;
const int maxn = 1e5+9;
const double esp = 1e-6;
char s[maxn];
int cnt ;
vector<int>v;
void solve(){
    vector<int>v;
    int sum = 0;
    scanf("%s" , s+1);
    int n = strlen(s+1);
    int ans = 1 ;
    cout << "Case #" << ++cnt << ": " ;
    rep(i , 2 , n){
        if(s[i] == s[i-1]){
            ans++;
        }else{
            sum += ans * ans ;
            v.pb(ans);
            ans = 1 ;
        }
    }
    v.pb(ans);
    sum += ans * ans ;
    int num = sum ;
    if(size(v) == 1){
        cout << sum << endl;
        return ;
    }
    for(int i = 0 ; i < size(v)-1 ; i++){
        if(i != size(v)-1 && i != 0 && v[i] == 1){
            int a = v[i-1]*v[i-1] + v[i]*v[i] + v[i+1]*v[i+1] , b = (v[i-1]+v[i]+v[i+1])*(v[i-1]+v[i]+v[i+1]);
            sum =max(sum , num - a + b) ;
        }else{
            int a = (v[i]*v[i] + v[i+1]*v[i+1]) , b = ((v[i]-1)*(v[i]-1))+((v[i+1]+1)*(v[i+1]+1));
            sum =max(sum , num - a + b) ;
            b = ((v[i]+1)*(v[i]+1))+((v[i+1]-1)*(v[i+1]-1));
            sum =max(sum , num - a + b) ;
        }
    }
    cout << sum << endl;
}

signed main()
{
    int t;
    scanf("%lld" , &t);
    while(t--)
        solve();
}